What is inverse oflaplace (s+3)/(s^2+4s+4) ?

The answer to inverse oflaplace (s+3)/(s^2+4s+4) is e^{-2t}+e^{-2t}t

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inverse oflaplace (s+3)/(s^2+4s+4)

inverse laplace

\frac{s + 3}{s ^{2} + 4 s + 4}

e^{- 2 t} + e^{- 2 t} t

Solution steps

L^{- 1} {\frac{s + 3}{s ^{2} + 4 s + 4}}

Take the partial fraction of

\frac{s + 3}{s ^{2} + 4 s + 4} : \frac{1}{s + 2} + \frac{1}{( s + 2 ) ^{2}}

= L^{- 1} {\frac{1}{s + 2} + \frac{1}{( s + 2 ) ^{2}}}

Use the linearity property of Inverse Laplace Transform:

For functions

f (s), g (s)

and constants

a, b : L^{- 1} {a \cdot f (s) + b \cdot g (s)} = a \cdot L^{- 1} {f (s)} + b \cdot L^{- 1} {g (s)}

= L^{- 1} {\frac{1}{s + 2}} + L^{- 1} {\frac{1}{( s + 2 ) ^{2}}}

L^{- 1} {\frac{1}{s + 2}} : e^{- 2 t}

L^{- 1} {\frac{1}{( s + 2 ) ^{2}}} : e^{- 2 t} t

= e^{- 2 t} + e^{- 2 t} t

\frac{1}{1 + 2 x ^{3}}

\int sec^{14} (x) tan (x) d x

\frac{d}{d x} (\frac{12}{x ^{5}})

\int sin (t) d t

\frac{d y}{d t} = y (8 - y)

What is inverse oflaplace (s+3)/(s^2+4s+4) ?
The answer to inverse oflaplace (s+3)/(s^2+4s+4) is e^{-2t}+e^{-2t}t